---
title: "500 kVA transformer efficiency, full and half load"
subject: "Electrical Machines"
chapter: 2
key: "semester_4/electrical_machines:tf-500kva-6600-400-efficiency"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-500kva-6600-400-efficiency
site: BCT Engineering
---

# 500 kVA transformer efficiency, full and half load

Electrical Machines · Chapter 2: Transformer

- Asked in: 2072 Asoj
- Video: [15. NUMERICAL PROBLEMS ON TRANSFORMER EFFICIENCY WITH DATA COPPER LOSS,  IRON LOSS, KVA, AND PF](https://www.youtube.com/watch?v=VM1cVkvdap4) — Electrical Tutorial
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-500kva-6600-400-efficiency
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2

## Question

A 500 kVA, 50 Hz, 6600 V/400 V, 1-phase transformer has primary and secondary winding resistances $0.4\ \Omega$ and $0.001\ \Omega$ respectively. If the iron loss is 3.0 kW, calculate the efficiency at (a) full load (b) half full load.

## Short answer

Full load copper loss is 3858.2 W, iron loss 3 kW. Efficiency at full load is 98.65% and at half load is 98.44%, both at unity power factor.

## Answer

Efficiency is output divided by input. Output = $x \times$ kVA $\times$ p.f., and losses = iron loss + $x^2 \times$ full-load copper loss, where $x$ is the fraction of full load.

The power factor is not given, so unity is assumed first (values for 0.8 lagging are added at the end).

#### Full-load currents

$$
I_1 = \frac{500000}{6600} = 75.76\ \text{A}, \qquad I_2 = \frac{500000}{400} = 1250\ \text{A}
$$

#### Full-load copper loss

$$
P_{cu} = I_1^2R_1 + I_2^2R_2 = (75.76)^2(0.4) + (1250)^2(0.001) = 2295.7 + 1562.5 = 3858.2\ \text{W}
$$

Iron loss $P_i = 3000$ W (constant at all loads).

#### (a) Full load (unity p.f.)

$$
\eta = \frac{500000}{500000 + 3000 + 3858.2} \times 100 = \frac{500000}{506858.2}\times 100 = 98.65\%
$$

#### (b) Half load (unity p.f.)

Copper loss at half load $= (0.5)^2 \times 3858.2 = 964.5$ W.

$$
\eta = \frac{250000}{250000 + 3000 + 964.5} \times 100 = \frac{250000}{253964.5}\times 100 = 98.44\%
$$

**Answer:** efficiency at full load $= 98.65\%$ and at half load $= 98.44\%$ (unity p.f.). At 0.8 p.f. lagging the values are 98.31% (full load) and 98.06% (half load).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Full-load copper loss is found as
   A. I1^2 R1 only
   B. I1^2 R1 + I2^2 R2 = 2295.7 + 1562.5 = 3858 W
   C. I1 R1 + I2 R2
   D. The same as the iron loss, 3000 W
   Answer: B. Both windings dissipate I^2R at their own rated currents (75.76 A and 1250 A). (Key point: Full-load copper loss from both windings)

2. When going from full load to half load, which losses change?
   A. Iron loss falls to a quarter, copper loss constant
   B. Both fall to a quarter
   C. Iron loss constant, copper loss falls to a quarter
   D. Both are constant
   Answer: C. Iron loss depends on voltage and is constant; copper loss varies as x^2 = 0.25, so 964.5 W. (Key point: Variation of losses with load)

3. At full load, unity pf, the efficiency is
   A. 98.44%
   B. 99.4%
   C. 97.5%
   D. 98.65%
   Answer: D. eta = 500000/(500000 + 3000 + 3858.2) = 98.65%. (Key point: Full-load efficiency)

4. The answer assumed unity pf. At 0.8 lagging the efficiency is lower because
   A. Losses increase at lower pf
   B. Output power (kVA x pf) is smaller while losses stay the same
   C. Core loss rises with lower pf
   D. Copper loss becomes zero
   Answer: B. Output falls to 400 kW while the losses are unchanged, so efficiency drops to 98.31%. (Key point: Effect of power factor)

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